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This question belongs to Arithmetic Ability Trigonometry
Trigonometry
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If cos(A - B) = $$\frac{{\sqrt 3 }}{2}$$ and sec A = 2, 0° ≤ A ≤ 90°, 0° ≤ B ≤ 90° then what is the measure of B?

Answer & Solution
Correct Answer: Option C
$$\eqalign{ & \cos \left( {A - B} \right) = \frac{{\sqrt 3 }}{2} \cr & \cos \left( {A - B} \right) = \cos {30^ \circ } \cr & A - B = {30^ \circ }........\left( {\text{i}} \right) \cr & \sec A = 2 \cr & \cos A = \frac{1}{2} = \cos {60^ \circ } \cr & A = {60^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & A = {60^ \circ }\,\& \,B = {30^ \circ } \cr} $$
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